Let \(A, B \in \mathcal {A}\)  be such that \(x \in A\)  and \(y \in B\) . There is a homeomorphic copy \(\sigma ^{\prime }\) 
of \(\sigma \)  which is the concatenation of an infinite genus half-strip in \(\Sigma _A\) ,
a zero genus compact strip in \(L\) , and an infinite genus half-strip in \(\Sigma _B\) .
Let \(g\)  be a homeomorphism sending \(\sigma \)  to \(\sigma ^{\prime }\) . For \(\mathcal {F}\)  and \(\nu \)  as in
Theorem REF , let \(\mathcal {F}^{\prime } = g^{-1} \mathcal {F}g\)  and \(\nu ^{\prime } = g^{-1} \nu g\) . Then
\(h_\sigma ^n \in (\mathcal {F}\nu )^k\qquad \Longleftrightarrow \qquad h_{\sigma ^{\prime }}^n \in (\mathcal {F}^{\prime } \nu ^{\prime })^k.\) 
That is, \(\langle h_\sigma \rangle \)  is CB if and only if \(\langle h_{\sigma ^{\prime }} \rangle \)  is CB.
In fact, in general, any set conjugate to a CB set is CB. Hence, it is enough to prove the Proposition for \(h_{\sigma ^{\prime }}\) .
Claim:
Assume there is no decomposition \(E^G = X \sqcup Y\)  where
\(x\in X\) , \(y \in Y\)  and \(X,Y\)  are closed \(\operatorname{FMap} (\Sigma )\) –invariant sets.
Then, there is a sequence
\(A = A_0, A_1, \dots , A_k = B, \qquad A_i \in \mathcal {A}\) 
and a sequence of ends \(z_i \in E^G\)  and \(z_i^{\prime } \in E(z_i)\) , \(i = 0, \dots , (k-1)\)  such that
\(z_0 \in A_0-\mathcal {M}(A_0)\) , \(z_{k-1}^{\prime } \in A_k-\mathcal {M}(A_k)\)  and, for \(i = 1, \dots , (k-1)\)  ,
\(z_{i-1}^{\prime }, z_i \in A_i-\mathcal {M}(A_{i-1})\) .
To see that this claim holds, let \(\mathcal {A}^{\prime }\)  be the subset of \(\mathcal {A}\)  such that, for \(C \in \mathcal {A}^{\prime }\) , there is a sequence
as described above starting from \(A\)  and ending in \(C\) . Let \(\mathcal {B}^{\prime } = \mathcal {A}- \mathcal {A}^{\prime }\) .
Then, for every \(A^{\prime } \in \mathcal {A}^{\prime }\) , \(B^{\prime } \in \mathcal {B}^{\prime }\)  and \(z \in E_{A^{\prime }}\cap E^G\)  we have
\(E(z) \cap E_{B^{\prime }} = \emptyset \) . Define
\(X = \bigcup _{A^{\prime } \in \mathcal {A}^{\prime }} E_{A^{\prime }} \cap E^G\qquad \text{and}\qquad Y = \bigcup _{B^{\prime } \in \mathcal {B}^{\prime } } E_{B^{\prime }}\cap E^G.\) 
The sets \(X\)  and \(Y\)  are closed and \(\operatorname{FMap} (\Sigma )\)  invariant, and the definition of \(X\)  implies that
we have \(x \in X\) . The assumption implies that \(y\)  cannot be in \(Y\) . Therefore \(B \in \mathcal {A}^{\prime }\)  and the
above sequence exists, which proves the claim.
<FIGURE>For \(i=0, \dots , (k-1)\) , fix disjoint sub-surfaces \(Z_i \subset \Sigma _{A_i} -x_{A_i}\)  and
\(Z_i^{\prime }\subset \Sigma _{A_{i+1}}-x_{A_{i+1}}\)  such that \(z_i\)  is an end of \(Z_i\) , \(z_i^{\prime }\)  is an end of \(Z_i^{\prime }\)  and
\(Z_i\)  is homeomorphic to \(Z_i^{\prime }\) , for example, we can choose \(Z_i\)  so that \(E(Z_i)\)  is an stable neighborhood
of \(z_i\) . Connect \(Z_i\)  to \(Z_i^{\prime }\)  by an arc \(\omega _i\)  that is contained in
\(L \cup \Sigma _{A_i} \cup \Sigma _{A_{i+1}}\) . Let \(Y_i\)  be a regular neighborhood of
\(Z_i \cup Z_i^{\prime } \cup \omega _i\) . Further, we assume that subsurfaces \(Y_i\)  are disjoint from each other
and are disjoint from the strip \(\sigma ^{\prime }\)  (see Figure REF ). For \(i =1, \dots (k-1)\) , let \(\partial _i\) 
be a sub-arc of \(\partial \Sigma _{A_i}\)  connecting \(\omega _{i-1} \cap \partial \Sigma _{A_i}\)  to
\(\omega _{i} \cap \partial \Sigma _{A_i}\) . Also, let \(\omega _i|_L\)  be the restriction
of \(\omega _i\)  to \(L\) . We further assume that the concatenation of \(\omega _i|_L\)  and \(\partial _i\) ,
which is an arc connecting \(\partial \Sigma _A\)  to \(\partial \Sigma _B\) , is homotopic (relative
\(\partial \Sigma _A \cup \partial \Sigma _B\) ) to either arc in the the restriction of \(\partial \sigma ^{\prime }\)  to
\(L\) , which we denote by \(\partial \sigma ^{\prime }|_L\) . Let \(\partial _0\)  be the sub-arc of \(\partial \Sigma _A\) 
connecting a point in \(\partial \sigma ^{\prime } \cap \partial \Sigma _A\)  to the points \(\omega _0 \cap \partial \Sigma _A\) 
and similarly let \(\partial _k\)  be the sub-arc of \(\partial \Sigma _B\)  connecting a point in
\(\partial \sigma ^{\prime } \cap \partial \Sigma _B\)  to the points \(\omega _{k-1} \cap \partial \Sigma _B\) .
Again, these can be chosen such that the concatenation \((\partial _i \omega _i|_L )_{i=0}^{(k-1)}\) 
is homotopic to the concatenation of an arc in \(\partial \sigma ^{\prime }|_L\)  and \(\partial _k\)  relative their end points.
Let \(g_i\)  be a homeomorphism that sends \(Z_i\)  to \(Z_i^{\prime }\) , whose support
is in \(Y_i\)  and that sends \(\omega _i\)  to itself in the reverse direction.
Let \(\mathcal {F}= \lbrace g_i\rbrace _{i=0}^{(k-1)}\) .
We now show that, for every \(n \in \mathbb {Z}\) , \(h_{\sigma ^{\prime }}^n \in (\nu _L \mathcal {F})^{k+1}\) . Let \(\sigma ^{\prime }_n\)  be
a sub-strip of \(\sigma ^{\prime }\)  of genus \(n\)  so that the genus of what remains between \(\sigma _n^{\prime }\)  and
\(\partial A\)  is zero. Choose an element \(f_0 \in \nu _{A_0}\)  that sends \(\sigma ^{\prime }_n\)  to a
\(Z_0\) , which is possible since \(z_0 \in E^G\)  and therefore \(Z_0\)  has infinite genus. In fact, we can do this in a way such that
\(f_0(\sigma ^{\prime })\cap \Sigma _A\)  starts the same as \(\sigma ^{\prime }\cap \Sigma _A\)  away from a small neighborhood of
\(\partial \Sigma _A\) , then it follows \(\partial _0\)  (staying in a small neighborhood of \(\partial \Sigma _A\) )
to the intersection point of \(\partial \Sigma _A\)  and \(\omega _0 \cap \Sigma _A\) , then follows
\(\omega _0 \cap \Sigma _A\)  (staying in \(Y_0\) ) to \(Z_0\) , then comes back the same way along
\(\omega _0 \cap \Sigma _A\)  and \(\partial _0\)  and continues along \(\sigma ^{\prime }\)  into \(B\) 
(see the left picture in Figure REF ). Then, \(g_0 \circ f_0(\sigma _n^{\prime }) \subset Z_0^{\prime }\) .
<FIGURE>We then find a \(f_1 \in \nu _{A_1}\)  that sends \(g_0 \circ f_0(\sigma _n^{\prime })\)  to \(Z_1\) . We can do this in a way such that
\(f_1 \circ g_0 \circ f_0 (\sigma ^{\prime })\)  starts the same as \(\sigma ^{\prime }\)  inside \(\Sigma _A\)  (away from a small neighborhood
of \(\partial \Sigma _A\) ) then it follows \(\partial _0\) , then \(\omega _0|_L\) , then \(\partial _1\)  to the intersection
point of \(\partial \Sigma _{A_1}\)  and \(\omega _1\) , then it follows \(\omega _1 \cap \Sigma _{A_1}\) 
(staying in \(Y_1\) ) to \(Z_1\) , then comes back the same way along \(\omega _1 \cap \Sigma _{A_1}\) , \(\partial _1\) ,
\(\omega _0|_L\)  and \(\partial _0\) , and then follows \(\sigma ^{\prime }\)  into \(B\) . We then apply \(g_1\)  and we have
\(g_1 \circ f_1 \circ g_0 \circ f_0(\sigma _n^{\prime }) \subset Z_1^{\prime }\) .
Following the same argument, we can find \(f_i\) , \(i =0, \dots , (k-1)\)  such that
\(\left( \prod _{i=0}^{k-1} g_i \circ f_i\right) (\sigma _n^{\prime }) \subset Z_{k-1}^{\prime } \subset \Sigma _{A_k} = \Sigma _B.\) 
The strip \(\prod g_i \circ f_i(\sigma ^{\prime })\)  starts the same as \(\sigma ^{\prime }\)  in \(\Sigma _A\) , then
follows the concatenation of the arcs \(\omega _i|_L\)  and \(\partial _i\)  (which by assumption is homotopic to arcs
\(\partial \sigma ^{\prime }|_L\) ), then follows \(\omega _k\cap \Sigma _B\)  to \(Z_k\) , then back the same way to
\(\partial \Sigma _A \cap \sigma ^{\prime }\) , and then continues the same as \(\sigma ^{\prime }\)  (see the right picture in
Figure REF ).
The portion of \(\prod _{i=0}^{k-1} g_i \circ f_i (\sigma _n^{\prime })\)  traveling back from \(\partial \Sigma _B\) 
to \(\partial \Sigma _A\)  has genus zero and can be homotoped into \(\Sigma _B\) . The portion going forward from
\(\partial \Sigma _A\)  to \(\partial \Sigma _B\)  also has genus zero and can be homotoped to a neighborhood of
the concatenation of \(\partial \sigma ^{\prime }|_L\)  and \(\partial _k\) . Hence, there is \(f_k \in \nu _B\)  such that
\(f_k \circ \prod g_i \circ f_i (\sigma ^{\prime })\)  can be homotoped into \(\sigma ^{\prime }\)  and hence
\(f_k \circ \prod g_i \circ f_i \)  can be considered as a map with support in \(\sigma ^{\prime }\) .
Since, \(\sigma _n^{\prime }\)  has moved from \(A\)  to \(B\) , we in fact have \(f_k \circ \prod g_i \circ f_i= h_{\sigma ^{\prime }}^n\) .
That is \(h_{\sigma ^{\prime }}^n \in (\nu _L \mathcal {F})^{k+1}\) . This finishes the proof.
Proposition 7.3 
Assume \(\Sigma _i\)  is a surface with either zero or infinite genus and where \(E(\Sigma _i)\)  has a single maximal
point \(z\) . If \(h_\sigma \)  is essential, then \(E(z)\)  is two-sided giving the decomposition
\(\mathrm {Accu} (x) = X \sqcup Y\)  where \(x \in X\)  and \(y \in Y\) .

The proof is nearly identical to the proof of Proposition REF . We outline it here.
Let \(x\) , \(y\)  be the ends of \(\Sigma \)  associated to the strip \(\sigma \) . Let \(A, B \in \mathcal {A}\)  be
such that \(x \in A\)  and \(y \in B\) . First find \(g \in \operatorname{FMap} (\Sigma )\)  such that \(\sigma ^{\prime }=g(\sigma )\) 
starts in \(\Sigma _A\) , continues in \(L\) , and then enters \(\Sigma _B\) . It is enough to to show that
the group generated by \(h_{\sigma ^{\prime }} = g h_\sigma g^{-1}\)  is CB.
Note that \(x,y \in \mathrm {Accu} (z)\) . We can find a sequence
\(A = A_0, A_1, \dots , A_k = B, \qquad A_i \in \mathcal {A}\) 
and a sequence of ends \(z_i \in \mathrm {Accu} (z)\)  and \(z_i^{\prime } \in E(z_i)\) , \(i = 0, \dots , (k-1)\)  such that
\(z_0 \in A_0-\mathcal {M}(A_0)\) , \(z_{k-1}^{\prime } \in A_k-\mathcal {M}(A_k)\)  and, for \(i = 1, \dots , (k-1)\)  ,
\(z_{i-1}^{\prime }, z_i \in A_i-\mathcal {M}(A_{i-1})\) . A similar argument to the proof of Proposition REF 
shows that if such a sequence does not exist, then \(E(z)\)  is two-sided which would be a contradiction.
We then construct sub-surfaces \(Z_i\)  and \(Y_i\)  and maps \(g_i\)  as before, and set
\(\mathcal {F}= \lbrace  g_i\rbrace _{i=0}^{k-1}\) . Let \(\sigma _n^{\prime }\)  be a sub-strip of \(\sigma ^{\prime }\)  that contains
the \(n\)  sub-surfaces \(\Sigma _i\)  that are nearest to \(\partial \Sigma _A\) . As in the
proof of Proposition REF , we can choose \(f_i \in \nu _{A_i}\)  such that \(f_k \circ \prod (g_i \circ f_i)\) 
sends \(\sigma ^{\prime }\)  to the strip that is homotopic to \(\sigma ^{\prime }\)  and moves \(\sigma _n\)  from \(\Sigma _A\)  to
\(\Sigma _B\) . That is
\(h_{\sigma ^{\prime }}^n = f_k \circ \prod (g_i \circ f_i) \in \big (\nu _L \mathcal {F}\big )^{k+1}.\) 
This finishes the proof.
Proposition 7.4 
Assume \(\Sigma _i\)  is a surface with either zero or infinite genus, and where \(\mathcal {M}(E(\Sigma _i))\)  is a Cantor set.
Then \(h_\sigma \)  is not essential.

As before, after conjugation, we can assume \(\sigma \)  intersects only \(\Sigma _A \cup L \cup \Sigma _B\) , where
\(\Sigma _i \in \Sigma _A\)  for \(i \le 0\)  and \(\Sigma _i \in \Sigma _B\)  for \(i>0\) .
Consider a simple closed curve in the strip \(\sigma \)  separating \(\Sigma _i\)  and \(\Sigma _{i+1}\) 
from the rest of the strip. Denote this subsurface by \(\Sigma _{i, i+1}\) . If the set of maximal
ends in \(\Sigma _i\)  is a Cantor set, the end space of \(\Sigma _i\)  is homemorphic to the end
space of \(\Sigma _{i, i+1}\)  (see Figure REF ).
<FIGURE>Similarly, the subsurface \(\Sigma _{i, i+n}\) , which is a
subsurface
of \(\sigma \)  with one boundary component containing the sub-surfaces \(\Sigma _i, \Sigma _{i+1}, \dots ,\Sigma _{i+n}\) , is also homeomorphic to \(\Sigma _i\) . Hence, for every \(n\) , there is a homeomorphism
\(f_n \in \nu _A\)  with support in \(\sigma \)  such that \(\Sigma _i\)  is sent to \(\Sigma _{i+n}\)  for
\(i < -n\) , \(\Sigma _i\)  is sent to itself for \(i>0\) , and the surface \(\Sigma _{-n, 0}\)  is sent to
\(\Sigma _0\) . Similarly, there there is a homeomorphism
\(g_n \in \nu _A\)  with support in \(\sigma \)  such that \(\Sigma _i\)  is sent to \(\Sigma _{i+n}\)  for
\(i >1\) , \(\Sigma _i\)  is sent to itself for \(i\le 0\) , and the surface \(\Sigma _{1}\)  is sent to
\(\Sigma _{1,n+1}\) . Then \(h_\sigma ^n = g_n h_\sigma f_n\) . That is, we can collect the surfaces
\(\Sigma _{-n} \dots \Sigma _0\)  to the surface \(\Sigma _0\) , shift \(\sigma \)  once and the expand
these surfaces to \(\Sigma _1, \dots , \Sigma _{n+1}\) , which means we have written \(h_\sigma ^n\)  as a composition of 3
homeomorphisms. By letting \(\mathcal {F}= \lbrace  h_\sigma \rbrace \) , we have that
\(\langle h_\sigma \rangle \)  is contained in \((\nu _L \mathcal {F})^2\) . Therefore, by Theorem REF ,
\(\langle h_\sigma \rangle \)  is a CB subset of \(\operatorname{FMap} (\Sigma )\)  and \(h_\sigma \)  is not essential.
We are now in a position where we can prove Theorem REF , which states that \(\operatorname{Map} (\Sigma )\) 
contains an essential shift if and only if the endspace of \(\Sigma \)  is two-sided.
One direction is already proven by Proposition REF .
It remains to show that, if \(E\)  is not two-sided then there is no essential shift map.
Assume \(E^G\)  is not two-sided and there does not exist any \(z \in E\)  such that \(E(z)\)  is two-sided.
Let \(h_\sigma \)  be any shift map with support on a strip \(\sigma \) , containing the subsurfaces \(\Sigma _i\) ,
exiting towards \(x, y \in E\) . The end space \(E(\Sigma _i)\)  has finitely many maximal types each of which is either a
Cantor set or a finite set. Therefore we can find finitely many sub-surfaces \(\Sigma _i^j \subset \Sigma _i\) ,
\(j=0 , \dots , \ell \) , each with one boundary component such that:

The surfaces \(\Sigma _i^j\)  are disjoint. Furthermore, \(\Sigma - \Sigma _i^j\)  is a compact planar surface
which means \(E(\Sigma _i) = \sqcup _j E(\Sigma _i^j)\) .

For \(j>0\) , \(\Sigma _i^j\)  has zero genus or infinite genus. That is, if \(\Sigma _i\)  has finite genus
we include all the genus in the surface \(\Sigma _i^0\)  so every other subsurface of \(\Sigma _i\)  contains no genus. If
\(\Sigma _i\)  already has zero genus or infinite genus, then there is no need to choose \(\Sigma _i^0\) . We can choose
\(\Sigma _i^0\)  to be a disk or have \(j\)  range from 1 to \(\ell \) . Also, if \(E(\Sigma _i^j) \cap E^G =\emptyset \) 
then we make sure \(\Sigma _i^j\)  has genus zero.

For every \(j=0, \dots , \ell \) , the surfaces \(\Sigma _i^j\)  are homeomorphic for all \(i \in \mathbb {Z}\) .
That is we decompose each \(\Sigma _i\)  in the same way.

Now the strip \(\sigma \)  can be decomposed to parallel strips \(\sigma ^j\)  each containing sub-surfaces
\(\Sigma _i^j\) . Since \(h_{\sigma ^j}\)  have disjoint support, they commute and \(h_\sigma = \prod h_{\sigma ^j}\) .
That is, the group generated by \(h_{\sigma _j}\) , \(j=0, \dots , \ell \) , is an abelian group that contains
the group generated by \(h_\sigma \) . Hence, if each \(\langle h_{\sigma ^j} \rangle \)  is CB then
\(\langle h_{\sigma ^0, \dots , h_{\sigma ^\ell }} \rangle \)  is also CB and thus \(\langle h_{\sigma } \rangle \) 
is CB.
Since \(E^G\)  is not two-sided, by Proposition REF , \(h_{\sigma ^0}\)  is not essential and
\(\langle h_{\sigma ^0} \rangle \)  is CB. For \(j=1, \dots , \ell \) , if \(\mathcal {M}(E(\Sigma _i^j))\)  is a single
point then \(\langle h_{\sigma ^0} \rangle \)  is CB by Proposition REF . If \(\mathcal {M}(E(\Sigma _i^j))\) 
is a Cantor set, then \(\langle h_{\sigma ^0} \rangle \)  is CB by Proposition REF .
This finishes the proof.
Finally, we prove Theorem REF  which tells us that a shift map is essential
if and only if there is either a decomposition of the ends accumulated by genus, or there is a
decomposition of the accumulation set of a maximal point of the surface.
We begin by proving the forward direction.
Consider the decomposition of the shift map \(h_\sigma = \prod _j h_{\sigma ^j}\)  constructed in the
proof of Theorem REF . As mentioned before,
if \(h_\sigma \)  is essential then some \(h_{\sigma _j}\)  is essential. If \(h_{\sigma _0}\)  is essential, then since \(\Sigma _i^0\)  contains finite genus,
Proposition REF  implies the first bullet point in Theorem REF  holds. If \(j>0\)  and
\(\mathcal {M}(E(\Sigma _i^j))\)  is a single point, then by Proposition REF  the second bullet point in
Theorem REF  holds. Finally, \(\mathcal {M}(E(\Sigma _i^j))\)  cannot be a Cantor set
by Proposition REF , which proves the forward direction.
In the other direction, first suppose that \(E^G\)  is two-sided giving a decomposition \(E^G = X \sqcup Y\) 
where \(x \in X\)  and \(y \in Y\) , then the proof of Proposition REF 
shows that \(h_{\sigma ^0}\)  is essential. In fact, there is a length function \({{\mathchoice{\hspace{0.55542pt}\mbox{$\centerdot $}}{\hspace{0.0pt}}{1}{m}u}{\hspace{0.55542pt}\mbox{$\centerdot $}}\hspace{0.55542pt}}{\hspace{0.83328pt}\centerdot \hspace{0.83328pt}}{\hspace{0.83328pt}\centerdot \hspace{0.83328pt}}\)\(such that, for \) n>0\(, \) h0n = 2n\(. Since, in this case, \) ij\( have genus zerofor \) j>0\(, the shift maps \) hj\( act trivially on homology. Hence \) hn = 2n\(and hence \) h\( is essential.\) Similarly, if there exists some \(z \in \mathcal {M}(\Sigma _i)\)  such that \(E(z)\)  is two-sided giving a
decomposition \(\mathrm {Accu}(z)= X \sqcup Y\)  where \(x \in X\)  and \(y \in Y\) , then the proof of
Proposition REF 
shows that \(h_{\sigma ^j}\)  is essential for some \(j>0\) . In fact, there is a length function \({{\mathchoice{\hspace{0.55542pt}\mbox{$\centerdot $}}{\hspace{0.0pt}}{1}{m}u}{\hspace{0.55542pt}\mbox{$\centerdot $}}\hspace{0.55542pt}}{\hspace{0.83328pt}\centerdot \hspace{0.83328pt}}{\hspace{0.83328pt}\centerdot \hspace{0.83328pt}}\)\(such that, for \) n>0\(, \) h0n = n\(. It is possible that the \) ij\( are homeomorphicfor several different \) j\(. We can assume, after re-indexing, that \) i1, ..., i'\(all have same type of countable maximal end. Then \) hn = ' n\(and hence \) h\( is essential.\)
Two sources of non-trivial topology
In this section, we prove Theorem REF  which states that if \(\Sigma \)  does not
have a non-displaceable subsurface and \(\operatorname{FMap} (\Sigma )\)  does not have an essential shift then
\(\operatorname{Map} (\Sigma )\)  and \(\operatorname{FMap} (\Sigma )\)  are CB.
This implies that \(\operatorname{Map} (\Sigma )\)  and \(\operatorname{FMap} (\Sigma )\)  are quasi-isometric to a point and therefore do not have an interesting
geometry. To prove this theorem, we make use of the notion of an avenue surface first introduced in [1]}.
Recall that we always assume that \(\Sigma \)  is stable.
Definition 8.1 An avenue surface is a connected orientable surface \(\Sigma \)  which does not contain any
non-displaceable subsurface of finite type, and whose mapping class
group \(\operatorname{Map} (\Sigma )\)  is CB-generated but not CB.

That is, the only possible examples for surfaces that have no non-displaceable
subsurfaces, no essential shifts, and are not CB are the avenue surfaces.
A topological description of avenue surfaces was given in [1]}.
Lemma 8.2 (Lemmas 4.5 and 4.6 in [1]})  
Let \(\Sigma \)  be an avenue surface. Then \(\Sigma \)  has either 0 or infinite genus, and
\(\Sigma \)  has exactly two ends of maximal type, that is, \(\mathcal {M}(E)=\lbrace  x_1, x_2\rbrace \) .
Furthermore, for every \(x \in E-\lbrace x_1, x_2\rbrace \) , the set \(E(x)\)  accumulates to both \(x_1\) 
and \(x_2\) .

In order to state the classification of CB mapping class groups from [4]}, we must first
define the notion of self-similarity for a space of ends, and the notion of telescoping for
an infinite-type surface.
Definition 8.3 
A space of ends \((E, E^G)\)  is said to be self-similar if for any decomposition
\(E = E_1 \sqcup E_2 \sqcup \ldots \sqcup E_n\)  of \(E\)  into pairwise disjoint clopen sets,
there exists a clopen set \(D\)  in some \(E_i\)  such that \((D, D\cap E^G)\)  is homeomorphic to \((E, E^G)\) .
A surface \(\Sigma \)  is telescoping if there are ends \(x_1, x_2 \in E\)  and disjoint
clopen neighborhoods \(V_i\)  of \(x_i\)  in \(\Sigma \)  such that for all clopen neighborhoods
\(W_i \subset V_i\)  of \(x_i\) , there exist homeomorphisms \(f_i, h_i \in \operatorname{FMap} (\Sigma )\) , with
\(f_i(W_i) \supset (\Sigma - V_{3-i})\qquad h_i(W_i) = Vi,\qquad \text{and}\qquad h_i(V_{3-i}) = V_{3-i}.\) 

A classification of CB mapping class groups was given in [4]}.
Theorem 8.4 (Theorem 1.7 in [4]})  
The group \(\operatorname{Map} (\Sigma )\)  is CB if and only if \(\Sigma \)  has infinite or zero genus and \(E\) 
is either self-similar or telescoping.

We are now ready to prove Theorem REF  restated below.
Theorem REF . (Two sources of non-trivial geometry).

If \(\operatorname{Map} (\Sigma )\)  does not have an essential shift and \(\Sigma \)  does not contain
a non-displaceable subsurface then \(\operatorname{Map} (\Sigma )\)  is quasi-isometric to a point.
By way of contradiction, suppose that \(\Sigma \)  does not have a non-displaceable subsurface, \(\operatorname{FMap} (\Sigma )\) 
does not contain an essential shift, and that \(\operatorname{FMap} (\Sigma )\)  is not CB.
Then by assumption, \(\Sigma \)  is an avenue surface.
By Lemma REF ,
\(\mathcal {M}(E)=\lbrace x_1, x_2\rbrace \)  and for any other \(x \in E - \lbrace x_1, x_2\rbrace \) , \(E(z)\)  accumulates to
both \(x_1\)  and \(x_2\) . Consider an end \(z \in \mathcal {M}(E-\lbrace x_1, x_2\rbrace )\) , that is, an end that is
maximal in \(E-\lbrace x_1, x_2\rbrace \) . If \(E(z)\)  is countable, then \(E\)  is two-sided with
\(X=\lbrace x_1\rbrace \)  and \(Y=\lbrace x_2\rbrace \) . This cannot happen since we are
assuming there are no essential shifts (see Theorem REF ).
Therefore, \(E(z)\)  is a Cantor set.
Furthermore, either \(\Sigma \)  has genus zero or it is infinite genus
and \(\lbrace x_1, x_2 \rbrace  \subset E^G\)  (otherwise, \(\operatorname{Map} (\Sigma )\)  is not even locally CB, see
[4]}). We now show that these assumptions imply
that \(\Sigma \)  is telescoping and, by Theorem REF , \(\operatorname{Map} (\Sigma )\)  is CB, which will prove our claim.
We check the definition of telescoping. For \(i=1,2\) , let \(V_i\)  be disjoint stable neighborhoods
of \(x_i\)  such that \(E-(V_1 \cup V_2)\)  contains every maximal type in \(E-\lbrace x_1, x_2\rbrace \) .
Let \(W_i\)  be the given smaller neighborhoods of \(x_i\) . Since the \(V_i\)  are stable neighborhoods,
we have that \(V_i\)  is homeomorphic to \(W_i\) . We claim that \(E-(V_1 \cup V_2)\)  is homeomorphic to
\(E-(V_1 \cup W_2)\) . Since all maximal types in \(E-\lbrace x_1, x_2\rbrace \)  are present in \(E-(V_1 \cup V_2)\) ,
the sets \(E-(V_1 \cup V_2)\)  and \(E-(V_1 \cup W_2)\)  have the same maximal types. But all
these types are Cantor sets. Therefore, for every \(y \in V_2-W_2\)  there is a \(z \in E-(V_1 \cup V_2)\) 
such that \(E(y)\)  accumulates to \(z\) . By [4]}, there are small neighborhoods \(U_y\) 
of \(y\)  and \(U_z\)  of \(z\)  such that \(U_y \cup U_z\)  is homeomorphic to \(U_z\) . The neighborhoods
\(U_y\)  give a covering of \(V_2-W_2\) , hence there is a finite sub-covering. By making the neighborhoods
smaller, we can assume that they are disjoint. Since each neighborhood \(U_y\)  can be absorbed into \(U_z\) ,
the set \(V_2-W_2\)  can be absorbed into \(E-(V_1 \cup V_2)\)  and hence, \(E-(V_1 \cup V_2)\)  is
homeomorphic to \(E-(V_1 \cup W_2)\) .
Now consider surfaces \(\Sigma _{V_i}\)  and \(\Sigma _{W_i}\) 
whose end points are \(V_i\)  and \(W_i\) . We can choose these surfaces so that \(\Sigma _{W_i} \subset \Sigma _{V_i}\) 
and so that \(\Sigma _{V_i}\)  is disjoint from \(\Sigma _{V_{3-i}}\) . Also, we can assume
\(\Sigma _{V_i} - \Sigma _{W_i}\)  and \(\Sigma -(\Sigma _{W_1} \cup \Sigma _{W_2})\)  all have
genus zero or infinity. The end space of \(\Sigma _{V_i}- \Sigma _{W_i}\) 
is homeomorphic to the end space of \(\Sigma -(\Sigma _{W_1} \cup \Sigma _{W_2})\) 
and they have the same
genus. Also \(\Sigma _{V_1}\)  and \(\Sigma _{W_1}\)  have homeomorphic end spaces and the same
genus. Hence, there is a map \(h_1\)  such that \(h_1(\Sigma _{W_1})=\Sigma _{V_1}\)  and
\(h_1(\Sigma _{V_2}) = h_1(\Sigma _{V_2})\) . The map \(h_2\)  in Definition REF  can
also be similarly constructed.
The construction of the maps \(f_i\)  follows similarly. As above, we can send
\(\Sigma _{W_1}\)  to \(\Sigma -\Sigma _{V_2}\)  while fixing some neighborhood
\(W_2^{\prime }\)  of \(x_2\)  as long as \(W_2^{\prime }\)  is small enough so that \(V_2-W_2^{\prime }\)  intersect \(E(z)\)  for every
\(z \in \mathcal {M}(E-\lbrace x_1, x_2\rbrace )\) . This proves that \(\Sigma \) 
is telescoping and hence \(\operatorname{Map} (\Sigma )\)  is CB. Note that since
\(\operatorname{FMap} (\Sigma )\)  and \(\operatorname{Map} (\Sigma )\)  are quasi-isometric this means
\(\operatorname{FMap} (\Sigma )\)  is also CB which proves the theorem.
